The block diagram shows the main steps in the balanced process for the production of
vinyl chloride from ethylene. Each block represents a reactor and several other processing
units. The main reactions are:
Block A, chlorination
$C_{2}H_{4}+Cl_{2} \longrightarrow C_{2}H_{4}Cl_{2}$ ,yield on ethylene 98 per cent

Question

The block diagram shows the main steps in the balanced process for the production of
vinyl chloride from ethylene. Each block represents a reactor and several other processing
units. The main reactions are:
Block A, chlorination

C_{2}H_{4}+Cl_{2} \longrightarrow C_{2}H_{4}Cl_{2}

,yield on ethylene 98 per cent

The block diagram shows the main steps in the balanced process for the production of
vinyl chloride from ethylene. Each block represents a reactor and several other processing
units. The main reactions are:
Block A, chlorination
[latex] C_{2}H_{4}+Cl_{2} \longrightarrow C_{2}H_{4}Cl_{2} [/latex] ,yield on ethylene 98 per cent

Block B, oxyhydrochlorination
[latex] C_{2}H_{4}+2HCl+\frac{1}{2} O_{2} \longrightarrow C_{2}H_{4}Cl_{2}+H_{2}O [/latex] ,yields: on ethylene 95 per cent,
on HCl 90 per cent

Block C, pyrolysis
[latex] C_{2}H_{4}Cl_{2}\longrightarrow C_{2}H_{3}Cl+HCl [/latex] yields: on DCE 99 per cent, on HCl 99.5 per cent

The HCl from the pyrolysis step is recycled to the oxyhydrochlorination step. The flow
of ethylene to the chlorination and oxyhydrochlorination reactors is adjusted so that the
production of HCl is in balance with the requirement. The conversion in the pyrolysis
reactor is limited to 55 per cent, and the unreacted dichloroethane (DCE) separated and
recycled.

Using the yield figures given, and neglecting any other losses, calculate the flow of
ethylene to each reactor and the flow of DCE to the pyrolysis reactor, for a production
rate of 12,500 kg/h vinyl chloride (VC).

Accepted Answer

Molecular weights: vinyl chloride 62.5, DCE 99.0, HCl 36.5.
VC per hour [latex] =\frac{12,500}{62.5} =200[/latex] kmol/h

Draw a system boundary round each block, enclosing the DCE recycle within the
boundary of step C.

Let flow of ethylene to block A be X and to block B be Y, and the HCl recycle be Z.
Then the total mols of DCE produced =0.98X + 0.95Y, allowing for the yields, and
the mols of HCl produced in block C

[latex]=\left(0.98X +0.95Y ight) 0.995=Z [/latex] (a)

Consider the flows to and product from block B

The yield of DCE based on HCl is 90 per cent, so the mols of DCE produced
[latex]=\frac{0.90Z}{2} [/latex]

Note: the stoichiometric factor is 2 (2 mol HCl per mol DCE).
The yield of DCE based on ethylene is 95 per cent, so
[latex]\frac{0.90Z}{2} =0.95Y [/latex]
[latex]Z=\frac{0.95 imes 2Y}{0.9}[/latex]
Substituting for Z into equation (a) gives
[latex] Y=\left(0.98X+0.95Y ight)0.995 imes \frac{0.9}{2 imes 0.95} [/latex]
Y = 0.837X (b)

Total VC produced = [latex]0.99 imes total [/latex] DCE, so
[latex]0.99\left(0.98X+0.95Y ight) =200[/latex] kmol/h
Substituting for Y from equation (b) gives X= [latex]\underline{\underline{113.8} }[/latex] kmol/h
and Y= [latex]0.837 imes 113.8=\underline{\underline{95.3} } [/latex] kmol/h

HCl recycle from equation (a)
Z = [latex]\left(0.98 imes 113.8+0.95 imes 95.3 ight)0.995=\underline{\underline{\underline{201.1} } }[/latex] kmol/h
Note: overall yield on ethylene =[latex]\frac{200}{\left(113.8+95.3 ight) } imes 100=\underline{\underline{96} }[/latex] per cent

Question 2.13: The block diagram shows the main steps in the balanced proce...

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